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Proof step 5 from proof rely-owns-none (previous step)

r ∈ rs0, ∀ r0. (g(r0) = g0(r) ∧ r0 ∈ rs ↔ g0(r0) = g0(r) ∧ r0 ∈ rs0) ∧ (g(r0) = g0(r) ∧ r0 ∈ rs → f(r0) = f0(r0)), ∀ r0. r0 ∈ rs → g(r0) ≠ g0(r) ⊦ # rs0 = 0, rs0 = ∅

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 6:
g(r) = g0(r) ∧ r ∈ rs ↔ g0(r) = g0(r) ∧ r ∈ rs0, r ∈ rs0, g(r) = g0(r) ∧ r ∈ rs → f(r) = f0(r), ∀ r0. r0 ∈ rs → g(r0) ≠ g0(r), ∀ r0. (g(r0) = g0(r) ∧ r0 ∈ rs ↔ g0(r0) = g0(r) ∧ r0 ∈ rs0) ∧ (g(r0) = g0(r) ∧ r0 ∈ rs → f(r0) = f0(r0)) ⊦ # rs0 = 0, rs0 = ∅

Rule argument

pos: left 2 terms: r

Additional information

Used terms: r